Problem: Simplify the following expression and state the condition under which the simplification is valid. $y = \dfrac{-2q^2 - 30q - 108}{3q^2 + 33q + 90}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ y = \dfrac {-2(q^2 + 15q + 54)} {3(q^2 + 11q + 30)} $ $ y = -\dfrac{2}{3} \cdot \dfrac{q^2 + 15q + 54}{q^2 + 11q + 30} $ Next factor the numerator and denominator. $ y = - \dfrac{2}{3} \cdot \dfrac{(q + 6)(q + 9)}{(q + 6)(q + 5)}$ Assuming $q \neq -6$ , we can cancel the $q + 6$ $ y = - \dfrac{2}{3} \cdot \dfrac{q + 9}{q + 5}$ Therefore: $ y = \dfrac{ -2(q + 9)}{ 3(q + 5)}$, $q \neq -6$